Provide an appropriate response. A survey of 300 non-fatal accidents showed that 108 involved uninsured drivers. Construct a 99% confidence interval for the proportion of fatal accidents that involved uninsured drivers. Round to the nearest thousandth. Answer: (L, U) = ( )
Provide an appropriate response. A survey of 300 non-fatal accidents showed that 108 involved uninsured drivers. Construct a 99% confidence interval for the proportion of fatal accidents that involved uninsured drivers. Round to the nearest thousandth. Answer: (L, U) = ( )
Calculus For The Life Sciences
2nd Edition
ISBN:9780321964038
Author:GREENWELL, Raymond N., RITCHEY, Nathan P., Lial, Margaret L.
Publisher:GREENWELL, Raymond N., RITCHEY, Nathan P., Lial, Margaret L.
Chapter13: Probability And Calculus
Section13.3: Special Probability Density Functions
Problem 7E
Related questions
Question
![---
**Confidence Interval for Proportion of Uninsured Drivers in Non-fatal Accidents**
Provide an appropriate response.
A survey of 300 non-fatal accidents showed that 108 involved uninsured drivers. Construct a 99% confidence interval for the proportion of fatal accidents that involved uninsured drivers. Round to the nearest thousandth.
Answer: (L, U) = ( [Text Box1] , [Text Box2] )
---
解释没有 graphs 或 diagrams，需要帮助客户理解给定信息。](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fd2297071-c636-4616-baf3-32dd51d4726a%2Fc76c7117-e67f-46ab-82cb-420f5a6d1b12%2F4lth9dn_processed.png&w=3840&q=75)
Transcribed Image Text:---
**Confidence Interval for Proportion of Uninsured Drivers in Non-fatal Accidents**
Provide an appropriate response.
A survey of 300 non-fatal accidents showed that 108 involved uninsured drivers. Construct a 99% confidence interval for the proportion of fatal accidents that involved uninsured drivers. Round to the nearest thousandth.
Answer: (L, U) = ( [Text Box1] , [Text Box2] )
---
解释没有 graphs 或 diagrams，需要帮助客户理解给定信息。
![### Constructing a 95% Confidence Interval for Proportion
**Provide an appropriate response.**
When 495 junior college students were surveyed, 145 said they have a passport.
Construct a 95% confidence interval for the proportion of junior college students that have a passport. Round to the nearest thousandth.
**Answer: (L, U) = ( [________], [________] )**
### Explanation:
To construct a 95% confidence interval for the proportion (\( \hat{p} \)) of junior college students that have a passport, follow these steps:
1. **Calculate the sample proportion** (\( \hat{p} \)):
\[
\hat{p} = \frac{145}{495}
\]
2. **Determine the standard error (SE)**:
\[
SE = \sqrt{\frac{\hat{p} (1 - \hat{p})}{n}}
\]
where \( n \) is the sample size.
3. **Find the critical value** for a 95% confidence level. The critical value (\( z* \)) can be found using a \( Z \)-distribution table or standard normal distribution, which is 1.96 for a 95% confidence level.
4. **Calculate the margin of error (ME)**:
\[
ME = z* \times SE
\]
5. **Construct the confidence interval (CI)**:
\[
CI = \hat{p} \pm ME
\]
Then, you would insert the numerical values into the answer box to complete the confidence interval.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fd2297071-c636-4616-baf3-32dd51d4726a%2Fc76c7117-e67f-46ab-82cb-420f5a6d1b12%2Fz8ei17_processed.png&w=3840&q=75)
Transcribed Image Text:### Constructing a 95% Confidence Interval for Proportion
**Provide an appropriate response.**
When 495 junior college students were surveyed, 145 said they have a passport.
Construct a 95% confidence interval for the proportion of junior college students that have a passport. Round to the nearest thousandth.
**Answer: (L, U) = ( [________], [________] )**
### Explanation:
To construct a 95% confidence interval for the proportion (\( \hat{p} \)) of junior college students that have a passport, follow these steps:
1. **Calculate the sample proportion** (\( \hat{p} \)):
\[
\hat{p} = \frac{145}{495}
\]
2. **Determine the standard error (SE)**:
\[
SE = \sqrt{\frac{\hat{p} (1 - \hat{p})}{n}}
\]
where \( n \) is the sample size.
3. **Find the critical value** for a 95% confidence level. The critical value (\( z* \)) can be found using a \( Z \)-distribution table or standard normal distribution, which is 1.96 for a 95% confidence level.
4. **Calculate the margin of error (ME)**:
\[
ME = z* \times SE
\]
5. **Construct the confidence interval (CI)**:
\[
CI = \hat{p} \pm ME
\]
Then, you would insert the numerical values into the answer box to complete the confidence interval.
Expert Solution
This question has been solved!
Explore an expertly crafted, step-by-step solution for a thorough understanding of key concepts.
This is a popular solution!
Trending now
This is a popular solution!
Step by step
Solved in 3 steps with 8 images
Recommended textbooks for you
Calculus For The Life Sciences
Calculus
ISBN:
9780321964038
Author:
GREENWELL, Raymond N., RITCHEY, Nathan P., Lial, Margaret L.
Publisher:
Pearson Addison Wesley,
Glencoe Algebra 1, Student Edition, 9780079039897…
Algebra
ISBN:
9780079039897
Author:
Carter
Publisher:
McGraw Hill
Holt Mcdougal Larson Pre-algebra: Student Edition…
Algebra
ISBN:
9780547587776
Author:
HOLT MCDOUGAL
Publisher:
HOLT MCDOUGAL
Calculus For The Life Sciences
Calculus
ISBN:
9780321964038
Author:
GREENWELL, Raymond N., RITCHEY, Nathan P., Lial, Margaret L.
Publisher:
Pearson Addison Wesley,
Glencoe Algebra 1, Student Edition, 9780079039897…
Algebra
ISBN:
9780079039897
Author:
Carter
Publisher:
McGraw Hill
Holt Mcdougal Larson Pre-algebra: Student Edition…
Algebra
ISBN:
9780547587776
Author:
HOLT MCDOUGAL
Publisher:
HOLT MCDOUGAL
Big Ideas Math A Bridge To Success Algebra 1: Stu…
Algebra
ISBN:
9781680331141
Author:
HOUGHTON MIFFLIN HARCOURT
Publisher:
Houghton Mifflin Harcourt